How to prove the concentration equality for standard normal?

The following inequality is given in some of Yale's online lecture notes

$$P(|Z|>x) \leq 2 \sqrt{2 \pi} \phi(x)$$

Where $$Z \sim N(0,1)$$ with density $$\phi(x)$$. They call it a concentration inequality, the proof is not given.

I know how to prove the bounds

$$(1/x - 1/x^3)\phi(x) \leq 1-\Phi(x) \leq 1/x \phi(x)$$

But the method I used for these did not get me anywhere.

I know that it is enough to show

$$\Phi(x) \geq 1- \sqrt{2 \pi} \phi(x)$$

by symmetry of normal distribution. So we need to show

$$\int_x^\infty \phi(y)dy \geq 1- \sqrt{2 \pi} \phi(x)$$

Edit:

Note that $$E[e^{Zt}] = e^{t^2/2}$$ and we already have the inequality

$$P(Z>x) \leq E[e^{Zt}]/e^{xt}$$

Maximising the RHS equality with respect to $$t$$ gives $$t_\max = x$$, and gives the upperbound needed.

It is not true. When $$x=0.1$$, $$P(|Z|>x) = 0.92$$ and $$2\phi(x)=0.79$$.

Maybe the correct one is:

$$P(|Z|>x) \leq 2 \sqrt{2\pi}\phi(x)$$

For new inequality, following follow step:

1. Get $$E(e^{\lambda Z})$$

2. Find $$\sup_\lambda(\lambda x - \log E(e^{\lambda Z}))$$

3. Use Chernoff bound, you can get $$\Pr(Z>x) < e^{-t^2/2}$$

• You're right. I forgot the normalisingn constant is not included. Will correct my post Dec 1, 2018 at 7:46
• Thanks a lot! Maximising the bound never occured to me but it's so obvious now :) Dec 1, 2018 at 11:21